Question: $\dfrac{d}{dx}\left(\dfrac{-2}{x^4}+\dfrac{1}{x^3}-x\right)=$
Explanation: The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}\dfrac{-2}{x^4}+\dfrac{1}{x^3}-x \\\\ &=-2x^{-4}+x^{-3}-x \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-2x^{-4}+x^{-3}-x) \\\\ &=-2\dfrac{d}{dx}(x^{-4})+\dfrac{d}{dx}(x^{-3})-\dfrac{d}{dx}(x) \\\\ &=-2(-4x^{-5})+(-3)x^{-4}-1x^0 \\\\ &=8x^{-5}-3x^{-4}-1 \\\\ &=\dfrac{8}{x^5}-\dfrac{3}{x^4}-1 \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{-2}{x^4}+\dfrac{1}{x^3}-x\right)=\dfrac{8}{x^5}-\dfrac{3}{x^4}-1$.